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horsena [70]
3 years ago
9

A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.

Physics
1 answer:
WINSTONCH [101]3 years ago
5 0

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

 sin∅ = 3.588 × 10⁻⁷

we know that;

 sin∅  = object separation / distance from telescope

object separation =   sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷  × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

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) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
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The frequency of the B note is

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The angular frequency instead is given by

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f = 466 Hz

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T = 50.0 ms = 0.050 s

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(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

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T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

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